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I have been trying to find an integral that wolfapha would not compute an answer and I have finaly found out.

My problem I don't know how to solve it.

$$\int \frac{\mathrm{d}x}{x+\sqrt{-x^2}}$$

Some help would be greatly liked.

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On the real line, what the software could do when the value under the square root was always negative? –  Babak S. Dec 20 '12 at 5:16
    
Regarding your initial question about finding a function which WA cannot integrate. Did you try, e.g., the function $ \exp(\sin x^2)$ –  Fabian Dec 20 '12 at 6:30
    
@BabakSorouh I really have no idea of exactly what you are saying, I am studying calculus. –  yiyi Dec 20 '12 at 7:22
    
@MaoYiyi: I was noting exactly what Ross pointed in a complete way below. ;-) –  Babak S. Dec 20 '12 at 7:25

2 Answers 2

up vote 2 down vote accepted

The most sensible interpretation of the problem I can find is to take $\sqrt {-x^2}$ as $\sqrt {(-x)^2}=|x|$ though I think the usual interpretation applies the $-$ after the square and would get $\sqrt{-(x^2)}$ and claim the square root is invalid. Accepting the first, you have $\int \frac {dx}{2x}$ which you can probably solve easily.

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Why not $\int \frac{1}{(1+i\, \text{sgn}\,x)} \frac{dx}{x}$? –  copper.hat Dec 20 '12 at 6:17
    
@copper.hat: also a nice choice. I think one should point out to the OP that this ambiguity is exactly the reason which hinders Mathematica from evaluating the integral... –  Fabian Dec 20 '12 at 6:19
    
@copper.hat: this looks like a real problem to me, mostly as the variable is $x$ instead of $z$. You are free to propose that solution. –  Ross Millikan Dec 20 '12 at 6:19
    
Then I think I would go with your invalid claim? –  copper.hat Dec 20 '12 at 6:22
    
@copper.hat: that is my approach. I think getting answers in both fields may help OP or others with this question. –  Ross Millikan Dec 20 '12 at 6:25

I would interpret the integral as either $\int \frac{1}{(1+i\, \text{sgn}\,x)} \frac{dx}{x}$, or invalid as in Ross' answer.

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